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I am currently doing research in Combinatorial Geometry and I have been able to reduce a quite complicated problem relating to extending the Newton-Gregory problem of kissing spheres to a simple number theory problem and then checking every case to see that a conjecture of mine holds.

In any case, for background reasons which are not necessary for me to get into, I need to determine the explicit sets of 4 positive integers which when summed together give 12. Order does not matter as I will need to permute the set of 4 positive integers in each case to satisfy a different case for verifying my conjecture by exhaustion. So, I am not interested in some abstract "there are this many ways", I am actually interested in generating the explicit sets of numbers. I have been able to come up with the following so far:

$$\{1,1,5,5\},\{2,2,4,4\},\{3,3,3,3\},\{2,2,3,5\},\{1,2,4,5\},\{1,3,3,5\},\{1,3,4,4\},\{2,3,3,4\}$$

Any ideas for how to solve this? I can clarify any ambiguities as needed!

EDIT: I forgot to mention the following important detail:

I only want to consider integers from the set $\{2,3,4,5\}$ in summing to 12, since these correspond to the degree of a vertex and I have proved that for my particular problem that all vertices have either degree 2, 3, 4, or 5.

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you may be interested in site.uottawa.ca/~ivan/F49-int-part.pdf –  Robert Mastragostino Jun 4 '12 at 19:26
    
If you will be permuting the four entries anyway, then you actually want to generate all possibilities with order (so that you want to also generate 1,5,1,5 and 5,5,1,1 as distinct from 1,1,5,5). In that case, the stars and bars method will give you not only the formula, but a method for producing the actual tuples. –  Arturo Magidin Jun 4 '12 at 19:26
    
@ArturoMagidin: I forgot to mention a very important detail which is that I only want to consider summing numbers from the set $\{2,3,4,5\}$. Does this change the stars and bars method? –  Samuel Reid Jun 4 '12 at 19:37
    
@SamuelReid: Yes. But you can simply generate all the possibilities and discard any with too-large entries. –  Arturo Magidin Jun 4 '12 at 19:40
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2 Answers 2

up vote 1 down vote accepted

You want $a+b+c+d = 12$ where $2 \leq a \leq b \leq c \leq d \leq 5$. Let $a_1 = a-2$, $b_1 = b-2$, $c_1 = c-2$ and $d_1 = d-2$. This gives us that $$a_1 + b_1 + c_1 + d_1 = 4$$ where $0 \leq a_1 \leq b_1 \leq c_1 \leq d_1 \leq 3$.

Let $b_1 = a_1 + b_2$, $c_1 = b_1 + c_2$ and $d_1 = c_1 + d_2$. Then we need $$a_1 + (a_1 + b_2) + (a_1 + b_2 + c_2) + (a_1 + b_2 + c_2 + d_2) = 4$$ i.e. $$4a_1 + 3b_2 + 2 c_2 + d_2 = 4$$ where $0 \leq a_1,b_2,c_2,d_2$.

Note that $a = a_1 +2$, $b = a_1 + b_2 + 2$, $c = a_1 + b_2 + c_2 + 2$ and $d = a_1 + b_2 + c_2 + d_2 + 2$

Now all we want is $$4a_1 + 3b_2 + 2 c_2 + d_2 = 4$$ such that $0 \leq a_1 \leq b_1 \leq c_1 \leq d_1 \leq 3$.

This means $a_1 \leq 1$.

If $a_1 = 1$, then $b_2 = c_2 = d_2 = 0$. Hence the solution is $$(a,b,c,d) = (3,3,3,3)$$

If $a_1 = 0$, then $$3b_2 + 2 c_2 + d_2 = 4$$ where $0 \leq b_2,c_2,d_2$.

This means $b_2 \leq 1$.

If $b_2 = 1$, then $c_2 = 0$ and $d_2 = 1$. Hence, the solution is $$(a,b,c,d) = (2,3,3,4)$$

If $b_2 = 0$, then $$2c_2 + d_2 = 4$$ where $0 \leq c_2,d_2$. This gives us $(c_2,d_2) = (2,0)$, $(c_2,d_2) = (1,2)$ and $(c_2,d_2) = (0,4)$. But $d_1 \leq 3$. Hence, the last solution is not possible.

Hence, these now give the solutions $$(a,b,c,d) = (2,2,4,4)$$ $$(a,b,c,d) = (2,2,3,5)$$

Hence, the only four possible solutions for $a+b+c+d = 12$, with the constraint that $a,b,c,d \in \{2,3,4,5\}$ are

$$(a,b,c,d) = (3,3,3,3)$$ $$(a,b,c,d) = (2,3,3,4)$$ $$(a,b,c,d) = (2,2,4,4)$$ $$(a,b,c,d) = (2,2,3,5)$$

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Except... you're not supposed to get any 6's, and you are missing things like (1,1,5,5)... –  Arturo Magidin Jun 4 '12 at 19:53
    
@ArturoMagidin I didn't respect the upper bound. Now I have changed it. $(1,1,5,5)$ is not acceptable. (The OP wants the numbers to be between $2$ and $5$). –  user17762 Jun 4 '12 at 19:56
    
@Mavis: Oh; since he listed (1,1,5,5) among his answers, I missed that. –  Arturo Magidin Jun 4 '12 at 19:57
    
@ArturoMagidin: I really confused myself while I was answering this question and it turns out that the question I asked is the problem I actually need to answer. In any case, with Marvis' response I was able to figure out the question which I did need to answer which I had not asked here. Sorry for the confusion, I appreciate the help. –  Samuel Reid Jun 4 '12 at 20:37
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Since order does not matter, assume $a \geq b \geq c \geq d \geq 1$ and $a + b + c + d = 12$. All you have to do is a case-by-case exhaustion of what possible values you can have. For example, the largest possible value of $a$ is $9$, in which case $b = c = d = 1$. If $a = 8 \dots$

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